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3: Summary Statistics
Notation
Consider these 10 ages (in years):
21
•
•
•
•
42
5
11
30
50
28
27
24
52
The symbol n represents the sample size (n = 10).
The capital letter X denotes the variable.
xi represents the ith value of variable X. For the data, x1 = 21, x2 = 42, and so on.
The symbol Σ (“capital sigma”) denotes the summation function. For the data, Σxi = 21 +
42 +…+ 52 = 290.
Measures of Central Location
Mean (arithmetic average)
The three main measures that summarize the center of a distribution are the mean, median, and
mode. While there are several different types of mean, we will focus on the arithmetic average.
To calculate the arithmetic mean, sum all the values and divide by n (equivalently, multiple 1/n):
x=
For the dataset, x =
1
∑ xi
n
1
(290) = 29 years.
10
Interpretation of the mean. The mean tells you:
•
•
•
•
The expected value of an individual drawn at random from the sample.
The expected value of an individual drawn at random from the population.
The expected value of the population mean.
The gravitational center of a distribution. This is where the distribution would balance:
The mean is like a seesaw. A small child can sit farther from the center of a seesaw in order to
balance an adult sitting closer to the center. Similarly, a single outlier sitting far off-center can have a
profound effect on the mean. The mean is sensitive to the effects of outliers.
�) and population mean (denoted µ) is critical
The distinction between the sample mean (denoted 𝒙𝒙
to your future understanding of statistical inferences.
(c) 2006; 2016 B. Gerstman
sumstats.docx
Page 1 of 7
Reporting the mean: The mean should be rounded to three (or at most four) significant digits
before reporting in order to avoid an appearance of pseudo-precision. In addition, units of measure
should be included when reporting the mean. For example, the mean of our data is 29.0 years.
Median
𝑛𝑛+1
The median was covered in the previous chapter. Briefly, this is the value with a depth of
in the
2
ordered array. When n is odds, this will fall on a value in the dataset. When n is even, this will fall
between two values. Here are the sizes of 5 benign tumors (cubic centimeters):
4
7
8
↑
median
11
12
The median has a depth of (5 + 1) / 2 = 3 and a value of 8.
The median is slightly more resistant to outliers. This means it resists the pull of outliers and is more
apt to stay put. For instance, if the biggest value in the above set had been mistakenly entered 120
instead of 12, the mean would shift from 8.4 to 30 but the median would stay put at 8.
Mode
The mode is the most frequently occurring value in a data set.
AGE
| Freq
%
------+--------------3
|
2
0.3%
4
|
9
1.4%
5
|
28
4.3%
6
|
37
5.7%
7
|
54
8.3%
8
|
85
13.0%
9
|
94
14.4%
10
|
81
12.4%
11
|
90
13.8%
12
|
57
8.7%
13
|
43
6.6%
14
|
25
3.8%
15
|
19
2.9%
16
|
13
2.0%
17
|
8
1.2%
18
|
6
0.9%
19
|
3
0.5%
------+--------------Total |
654 100.0%
← Mode (most frequent value)
The mode is unreliable in all but very large data sets.
(c) 2006; 2016 B. Gerstman
sumstats.docx
Page 2 of 7
Comparison of the Mean, Median, and Mode
The mean, median, and mode are equivalent when the distribution is unimodal and symmetrical.
However, with asymmetry, the median is approximately one-third the distance between the mean and
mode:
The mean, median, and mode offer different advantages and disadvantages. The mean offers the
advantages of familiarity and efficiency. It also has advantages when making inferences about a
population mean. On the downside, the mean is markedly influenced by extreme skewness and
outliers. Under circumstances of extreme skewness, the median is a more “stable.” An often cited
example of this advantage come when considering the salary of employees, where the salary of
highly paid executives skews the average income toward a misleadingly high value. Another
example is the average price of homes, in which case high priced homes skew the data in a positive
direction. In such circumstances, the median is less likely to be misinterpreted, and is therefore the
preferred measure of central location.
You can judge the asymmetry of a distribution by comparing its mean and median. When the mean
is greater than the median, the distribution has a positive skew. When the mean is about equal to the
median, the distribution is symmetrical. When the mean is less than the median, the distribution has a
negative skew:
mean > median : positive skew
mean ≅ median : symmetry
mean < median : negative skew
In general, the mean is preferred when data are symmetrical and do not have outliers. In other
instances, the median may be the preferred measure of central location.
(c) 2006; 2016 B. Gerstman
sumstats.docx
Page 3 of 7
Measures of Spread
Range
Simply noting the minimum and maximum values can be useful when describing the spread of a
distribution. However, calculating the sample range (maximum – minimum) is not an acceptable
measure of spread because it will consistently underestimate the population range.
5-point summaries and quartiles
The 5-point summary consists of:
Q0 ≡ Quartile 0 (the minimum)
Q1 ≡ Quartile 1 (bigger than 25% of the data points)
Q2 ≡ Quartile 2 (the median)
Q3 ≡ Quartile 3 (bigger than 75% of the data points)
Q4 ≡ Quartile 4 (the maximum)
The median is called Q2 because it is equal to or greater than 2-quarters of the data points. The
minimum is Q0 because it is greater than or equal to zero-quarters of the data points. You get it.
When the data set is large (n >= 100), it is easy to find Q1 and Q3. With small data sets, the exact
location of quartiles must be interpolated. The two most common methods of interpolation for this
purpose are weighted averages and Tukey’s hinges. To find Tukey’s hinges:
•
•
•
•
•
Put the data in rank order
Locate the median of the data set.
Divide the data set into two groups: a low group and a high group. When n is odd, the
median should be placed in both groups.
Find the ‘median’ of the low group. This is Q1.
Find the ‘median’ of the high group. This is Q3.
Example #1. Consider this ordered array (n = 10)
5
11
21
24
low group
27
28
M
30
42
50
high group
52
The low group has a ‘median’ of 21. This is Q1. The high group has a ‘median’ of 42. This is Q3.
The five-point summary (5, 21, 27.5, 42, 52)
Example #2. Consider this new ordered array (n = 7) .
1.47
2.06
2.36
low group
3.43
M
3.74
3.78
high group
3.94
Recall that you must include the median in both the low group and high group for Tukey’s hinges.
Therefore the low group consists of {1.47, 2.06, 2.36, 3.43}. The ‘median’ of this low group (Q1) is
the average of 2.06 and 2.36, or 2.21. The ‘median’ of the high group (Q3) is 3.76. The five point
summary is (1.47, 2.21, 3.43, 3.76, 3.95).
(c) 2006; 2016 B. Gerstman
sumstats.docx
Page 4 of 7
Inter-quartile range (IQR)
An good measure of spread (esp. for asymmetrical data) is the inter-quartile range (IQR):
IQR = Q3 − Q1
The IQR for Example #1 (prior page) = 42 – 21 = 21. For Example #2 (prior page), IQR = 3.76 –
2.21 = 1.55. Our IQRs are also called hinge-spreads because they quantify the spread from the
lower hinge to the upper hinge.
Box-and-whiskers plots (boxplots)
The Tukey boxplot consists of a box showing Q1, Q2, and Q3, whiskers and, occasionally outside
values.
After determining the 5 point summary and IQR for a dataset, then calculate (but do not draw)
fences as follows:
•
•
FenceLower = Q1 − (1.5)(IQR)
FenceUpper = Q3 + (1.5)(IQR)
Note that the fences are 1.5 hinge-spreads above below the hinges. Do not plot these fences. Any
value that is above the upper fence is an upper outside value. Any values below the lower fence is a
lower outside value. Plot these points, if any, on the graph.
The largest value still inside the upper fence is the upper inside value. The smallest value still inside
the lower fence is the lower inside value. Drawn whiskers from the upper hinge to the upper inside
value and from the bottom hinge to the lower inside value.
Boxplot example #1.
5
11
21
24
27
28
30
42
50
52
The 5-point summary (determine on prior page) is (5, 21, 27.5, 42, 52).
The box extends from 21 to 42 and has a line in its midst to identify the
median at 27.5.
The FL = 21 − (1.5)(21) = −10.5 and FU = 42 + (1.5)(21) = 73.5. No
values in the data set are above 73.5 or below −10.5.Therefore, there are
no outside values. The upper inside value is 52 and the lower inside is the
5. Whiskers are drawn from the hinges to the inside values. The boxplot is
shown on the next page.
(c) 2006; 2016 B. Gerstman
sumstats.docx
Page 5 of 7
Interpretation of boxplots. Think shape, location and spread.
•
•
•
Shape is easiest to consider in terms of symmetry and potential outliers. Is the median about
halfway between the hinges? Is the box about half-way between the whiskers? Outside
values are potential outliers.
The central location is be summarized by the median and box, which is the middle 50% of
values.
Spread is quantified in terms of the hinge-spread and whisker-spread.
The boxplot for example #1 is fairly symmetrical and has no outside values. The a median that is a
little less than 30 and a hinge-spread is from 21 to 42.
New example: Here’s a new set of values:
3
21
22
24
25
26
28
29
31
51
The five-point summary is (3, 22, 25.5, 29, 51). IQR = 29 − 22
= 7.
FU = 29 + (1.5)(7) = 39.5. FL = 22 − (1.5)(7) = 11.5. There is
one value above the upper fence (51). There is one value
below the lower fence (3). The largest value still inside the
upper fences (upper inside value) is 31.
The smallest value still inside the lower fence (lower inside
value) is 21.
(c) 2006; 2016 B. Gerstman
sumstats.docx
Page 6 of 7
Standard Deviation and Variance
Both the standard deviation and variance are based on deviations (di), defined as
di = 𝑥𝑥𝑖𝑖 − 𝑥𝑥̅
Square each deviation in the dataset and sum them to derive the sum of squares (SS):
SS = ∑ ( xi − x ) 2
The sample variance is the “mean” sum of squares: s 2 =
1
⋅ SS
n −1
Note: The denominator in this formula is by n − 1, not n. This is necessary to derive an unbiased
estimate of the population variance. The number n − 1 is called the degree of freedom.
The sample standard deviation is simply the square root of the variance: s =
s2
Illustrative example. Recall these 10 ages: 21, 42, 5, 11, 30, 50, 28, 27, 24, 52. The
sample mean x is 29.0 (page 1). The variance is calculated:
i
1
2
3
4
5
6
7
8
9
10
Sums
Values (xi)
21
42
5
11
30
50
28
27
24
52
290
Deviations (di)
21 – 29 = –8
42 – 29 = 13
5 – 29 = –24
11 – 29 = –18
30 – 29 = 1
50 – 29 = 21
28 – 29 = –1
27 – 29 = –2
24 – 29 = –5
52 – 29 = 23
0
Sqyared d2
–82 = 64
132 = 169
–242 = 576
−182 = 324
12 = 1
212 = 441
−12 = 1
−22 = 4
−52 = 25
232 = 529
Sum of Squares  2134
The sample variance is s 2 =
SS
2134
=
= 237.1111 years2. [Note the squared units.]
n − 1 10 − 1
The standard deviation s =
s 2 = 237.1111 = 15.398 = 15.4 years.
Report the standard deviation in conjunction with the mean, round accordingly, and include units of
measure. “The mean age of the participants was 29.0 years with a standard deviation of 15.4 years.”
Calculating a few variance and standard deviations by hand is instructive.
However, most of the time we calculate the standard deviation with a
computer or calculator.
Do not need to calculate variances and standard deviations by hand unless
the instructions specifically request a step-by-step calculation.
(c) 2006; 2016 B. Gerstman
sumstats.docx
Page 7 of 7